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\title{Theoretical Assignment II}

\author{韩骐骏 \\ (数学科学学院)信息与计算科学3200103585}

\begin{document}

\maketitle

\section{Problem 2.9.1 I}
\subsection{Question 1}
By substitution we can get that $f_{\mbox{\scriptsize 0}}=f(x_{\mbox{\scriptsize 0}})=1, f_{\mbox{\scriptsize 1}}=f(x_{\mbox{\scriptsize 1}})=\frac{1}{2}$, thus$$f(x)-p_{\mbox{\scriptsize 1}}(f;x)=\frac{1}{x}-(-\frac{1}{2}x+\frac{3}{2})=\frac{1}{\xi (x)^3}(x-1)(x-2)$$Thus $$\xi(x)=\sqrt[3]{2x}$$

\subsection{Question 2}
From $1$ we can compute that$$max \xi(x)=\sqrt[3]{4}, min \xi(x)=\sqrt[3]{2}, max f''(\xi(x))=max \frac{1}{x}=1$$

\section{Problem 2.9.1 II}
\subsection{Solution}
there exists $g(x)\in \mathbb{R}_{\mbox{\scriptsize n}}$, such that$$g(x_{\mbox{\scriptsize i}})=\sqrt{f_{\mbox{\scriptsize i}}},i=0,1,...,n$$Let $p(x)=g^2(x)\geq 0$, thus $$p\in \mathbb{R}_{\mbox{\scriptsize 2n}}^+,p(x_{\mbox{\scriptsize i}})=g^2(x_{\mbox{\scriptsize i}})=f_{\mbox{\scriptsize i}}$$

\section{Problem 2.9.1 III}
\subsection{Question 1}
We prove it with mathematical induction. Obviously $$n=0,f[t]=e^t=\frac{(e-1)^0}{0!}e^t$$If the conclusion is right for $n=k$ with every $t\in \mathbb{R}$, then when $n=k+1$, $$f[t,t+1,...,t+k+1]=\frac{f[t+1,...,t+k+1]-f[t,...,t+k]}{k+1}=\frac{(e-1)^k^+^1}{k!(k+1)}e^t$$To sum up, the conclusion is correct.\\
\subsection{Question 2}
With the conclusion from Question 1 we know that $$f[0,1,...,n]=\frac{(e-1)^n}{n!}=\frac{1}{n!}f^(^n^)(\xi)$$That is $$e^\xi=(e-1)^n$$ $$\xi=n\ln (e-1)\geq 0.54n\geq \frac{n}{2}$$So $\xi$ is located to the right of $\frac{n}{2}$.\\

\section{Problem 2.9.1 IV}
\subsection{Question 1}
With difference table we can compute that $$f[0]=5,f[0,1]=-2,f[0,1,2]=1,f[0,1,2,3]=\frac{1}{4}$$Thus $$p_{\mbox{\scriptsize 3}}(f;x)=5-2x+x(x-1)+\frac{1}{4}x(x-1)(x-3)=\frac{1}{4}x^3-\frac{9}{4}x+5$$
\subsection{Question 2}
From Question 1 we can find that$$p'_{\mbox{\scriptsize 3}}(f;x)=\frac{3}{4}x^2-\frac{9}{4},x_{\mbox{\scriptsize min}}=\sqrt{3}$$

\section{Problem 2.9.1 V}
\subsection{Solution}
\begin{tabular}{c|c|c|c|c|c|c}
    0 & 0 \\
    1 & 1   & 1 \\
    1 & 1   & 7   & 6 \\
    1 & 1   & 7   & 21  & 15\\
    2 & 128 & 127 & 120 & 99 & 42 \\
    2 & 128 & 448 & 321 & 201 & 102 & 30
\end{tabular}
With this difference table we can compute that $f[0,1,1,1,2,2]=30$, thus $$30=f[0,1,1,1,2,2]=\frac{f^(^5^)(\xi)}{5!}=21\xi^2,\xi=\sqrt{\frac{10}{7}}$$

\section{Problem 2.9.1 VI}
\subsection{Solution}
\begin{tabular}{c|c|c|c|c|c}
    0 & 1 \\
    1 & 2 & 1 \\
    1 & 2 & -1 & -2 \\
    3 & 0 & -1 & 0   & $\frac{2}{3}$ \\
    3 & 0 & 0  & $\frac{1}{2}$ & $\frac{1}{4}$ & -$\frac{5}{36}$
\end{tabular}
With this difference table we can compute that $$p_{\mbox{\scriptsize 4}}(x)=1+x-2x(x-1)+\frac{2}{3}2(x-1)^2-\frac{5}{36}x(x-1)^2(x-3)$$Thus $$f(2)\approx p_{\mbox{\scriptsize 4}}(2)=\frac{11}{18}$$ $$|f(2)-p_{\mbox{\scriptsize 4}}(2)|\leq \frac{|f^{(5)}(\xi)|}{5!}(2-1)^2(2-3)^2(2-0)\leq \frac{M}{60}$$

\section{Problem 2.9.1 VII}
\subsection{Solution}
It can be proved by induction, first we have $\Delta ^0f(x)=0!h^0f[x_{\mbox{\scriptsize 0}}]$, if we already have $$\Delta^kf(x)=k!h^kf[x_{\mbox{\scriptsize 0}},...,x_{\mbox{\scriptsize k}}]$$Then we will have $$\Delta^{k+1}f(x)=\Delta^kf(x+h)-\Delta^kf(x)=k!h^k(f[x_{\mbox{\scriptsize 1}},...,x_{\mbox{\scriptsize k+1}}]-f[x_{\mbox{\scriptsize 0}},...,x_{\mbox{\scriptsize k}}])=(k+1)!h^{k+1}f[x_{\mbox{\scriptsize 0}},\dots,x_{\mbox{\scriptsize k+1}}]$$The second case is the same.\\

\section{Problem 2.9.1 VIII}
\subsection{Solution}
$$\frac{\partial}{x_{\mbox{\scriptsize 0}}}f[x_{\mbox{\scriptsize 0}},...,x_{\mbox{\scriptsize n}}]=\lim_{h \rightarrow 0}\frac{f[x_{\mbox{\scriptsize 0}}+h,...,x_{\mbox{\scriptsize n}}]-f[x_{\mbox{\scriptsize 0}},...,x_{\mbox{\scriptsize n}}]}{h}$$ $$\frac{\partial}{x_{\mbox{\scriptsize 0}}}f[x_{\mbox{\scriptsize 0}},...,x_{\mbox{\scriptsize n}}]=\lim_{h \rightarrow 0}f[x_{\mbox{\scriptsize 0}},x_{\mbox{\scriptsize 0}}+h,...,x_{\mbox{\scriptsize n}}]=f[x_{\mbox{\scriptsize 0}},x_0,...,x_{\mbox{\scriptsize n}}]$$The partial derivative with respect to one of the other variables is similiar, $$\frac{\partial}{x_{\mbox{\scriptsize k}}}f[x_{\mbox{\scriptsize 0}},...,x_{\mbox{\scriptsize n}}]=f[x_{\mbox{\scriptsize k}},x_{\mbox{\scriptsize 0}},...,x_{\mbox{\scriptsize n}}]$$

\section{Problem 2.9.1 IX}
\subsection{Solution}
Actually, $$\min\max_{x\in[-1,1]}|a_{\mbox{\scriptsize 0}}x^n+...+a_{\mbox{\scriptsize n}}|=\frac{1}{2^{n-1}}$$Now we set $x=\frac{b-a}{2}t+\frac{a+b}{2}$, then $$x\in [a,b],t\in [-1,1]$$And $$\min\max_{x\in[a,b]}|a_{\mbox{\scriptsize 0}}x^n+...+a_{\mbox{\scriptsize n}}|=\min\max_{t\in[-1,1]}|a_{\mbox{\scriptsize 0}}(\frac{b-a}{2}t+\frac{a+b}{2})^n+...+a_{\mbox{\scriptsize n}}|$$ $$\min\max_{x\in[a,b]}|a_{\mbox{\scriptsize 0}}x^n+...+a_{\mbox{\scriptsize n}}|=a_{\mbox{\scriptsize 0}}(\frac{b-a}{2})^n\frac{1}{2^{n-1}}=\frac{a_{\mbox{\scriptsize 0}}(b-a)^n}{2^n}$$

\section{Problem 2.9.1 X}
\subsection{Solution}
Prove by disproof, if the conclusion is not true, thus $$\exists p_{\mbox{\scriptsize 0}} \in \mathbb{P}^a_{\mbox{\scriptsize n}},\Vert p_{\mbox{\scriptsize 0}} \Vert_\infty<\Vert \hat{p}_{\mbox{\scriptsize n}}\Vert_\infty$$Let $f=\hat{p}-p_{\mbox{\scriptsize 0}},f(x_{\mbox{\scriptsize k}}')=\frac{(-1)^k}{T_{\mbox{\scriptsize n}}(a)}-p_{\mbox{\scriptsize 0}}(x_{\mbox{\scriptsize k}}'),k=0,...,n$, because equation $|p_{\mbox{\scriptsize 0}}|<\frac{1}{T_{\mbox{\scriptsize n}}(a)}$ holds good under all circumstances, thus $f$ interpolates x-axis for at least n times. But the degree of $f$ is lower than $(n+1)$, thus $f=0$, which produces a contradiction.\\

\section{Problem 2.9.1 XI}
\subsection{Question a}
Obviously $b_{\mbox{\scriptsize n,k}}(t)=C^k_{\mbox{\scriptsize n}}t^k(1-t)^{n-k}>0$ when $t\in (-1,1)$.

\subsection{Question b}
$$\sum_{k=0}^n b_{\mbox{\scriptsize n,k}}(t)=\sum_{k=0}^n C^k_{\mbox{\scriptsize n}}t^k(1-t)^{n-k}=(t+1-t)^n=1$$

\subsection{Question c}
$kC^k_{\mbox{\scriptsize n}}=nC^{k-1}_{\mbox{\scriptsize n-1}}$, thus $$\sum_{k=0}^n kb_{\mbox{\scriptsize n,k}}(t)=\sum_{k=0}^n ntC^{k-1}_{\mbox{\scriptsize n-1}}t^{k-1}(1-t)^{n-k}=nt$$

\subsection{Question d}
$$\sum_{k=0}^n (k-nt)^2b_{\mbox{\scriptsize n,k}}(t)=\sum_{k=0}^n k^2b_{\mbox{\scriptsize n,k}}(t)-2nt\sum_{k=0}^n kb_{\mbox{\scriptsize n,k}}(t) + n^2t^2\sum_{k=0}^n b_{\mbox{\scriptsize n,k}}(t)$$ $kC^k_{\mbox{\scriptsize n}}=nC^{k-1}_{\mbox{\scriptsize n-1}}$,thus $$\sum_{k=0}^n k^2b_{\mbox{\scriptsize n,k}}(t)=nt\sum_{k=0}^n [(k-1)+1]C^{k-1}_{\mbox{\scriptsize n-1}}t^{k-1}(1-t)^{n-k}=nt[(n-1)t+1]=n^2t^2-nt^2+nt$$Combining b and c, we can get $$\sum_{k=0}^n (k-nt)^2b_{\mbox{\scriptsize n,k}}(t)=n^2t^2-nt^2+nt-2n^2t^2+n^2t^2=nt-nt^2$$

\end{document}
